Assume that the random variable Sn is the sum of 48 independent experimental values of the random variable X whose PDF is given by fX (x)= {1/3 1≤x≤4 0 otherwise Find the probability that Sn lies in the range 108≤Sn≤126.

Question

Assume that the random variable [latex]S_n[/latex] is the sum of 48 independent experimental values of the random variable [latex]X[/latex] whose PDF is given by

[latex]\,\displaystyle f_X\left (x ight )= \begin{cases}\frac{1}{3} & 1 \leq x \leq 4 \ 0 & ext { otherwise }\end{cases}[/latex]

Find the probability that [latex]S_n[/latex] lies in the range[latex]\,\displaystyle 108 \leq S_n \leq 126[/latex]

Accepted Answer

The expected value and variance of [latex]X[/latex] are given by

[latex]\,\displaystyle \,\displaystyle E[X]=\frac{4+1}{2}=\frac{5}{2} \[/latex]

[latex]\,\displaystyle \sigma_X^2=\frac{\left (4-1\right )^2}{12}=\frac{3}{4}\[/latex]

Also, [latex]S_n=X_1+X_2+...+X_{48}[/latex]. Because the [latex]X_i[/latex] are independent and identically distributed, the mean and variance of [latex]S_n[/latex] are given by

[latex]\,\displaystyle E\left[S_n\right]=48 E[X]=120\[/latex]

[latex]\,\displaystyle \sigma_{S_n}^2=48 \sigma_X^2=36[/latex]

Assuming that the sum approximates the normal random variable, which is usually true for [latex]n\geq 30[/latex], the CDF of the normalized random value of [latex]S_n[/latex] becomes

[latex]\,\displaystyle P\left[S_n \leq s\right]=F_{S_n}\left (s\right )=\Phi\left (\frac{s-E\left[S_n\right]}{\sigma_{S_n}}\right )=\Phi\left (\frac{s-120}{6}\right )[/latex]

Therefore, the probability that [latex]S_n[/latex] lies in the range [latex]\,\displaystyle 108 \leq S_n \leq 126[/latex] is given by

[latex]\,\displaystyle P\left[108 \leq S_n \leq 126\right]=F_{S_n}\left (126\right )-F_{S_n}\left (108\right )=\Phi\left (\frac{126-120}{6}\right )-\Phi\left (\frac{108-120}{6}\right )\[/latex]

[latex]\,\displaystyle =\Phi\left (1\right )-\Phi\left (-2\right )=\Phi\left (1\right )-\{1-\Phi\left (2\right )\}=\Phi\left (1\right )+\Phi\left (2\right )-1\[/latex]

[latex]\,\displaystyle =0.8413+0.9772-1=0.8185[/latex]

where the values of [latex]\Phi (1)[/latex] and [latex]\Phi (2)[/latex] are obtained from Table 1 in Appendix 1.

Table 1 Area Φ(x) under the standard normal curve to the left of x
x	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.5910	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.6480	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.6700	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.6950	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.7190	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.7580	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.7910	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078	0.8106	0.8133
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340	0.8365	0.8389
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577	0.8599	0.8621
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790	0.8810	0.8830
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944	0.8962	0.8980	0.8997	0.9015
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115	0.9131	0.9147	0.9162	0.9177
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265	0.9279	0.9292	0.9306	0.9319
1.5	0.9332	0.9345	0.9357	0.9370	0.9382	0.9394	0.9406	0.9418	0.9429	0.9441
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505	0.9515	0.9525	0.9535	0.9545
1.7	0.9554	0.9564	0.9573	0.9582	0.9591	0.9599	0.9608	0.9616	0.9625	0.9633
1.8	0.9641	0.9649	0.9656	0.9664	0.9671	0.9678	0.9686	0.9693	0.9699	0.9706
1.9	0.9713	0.9719	0.9726	0.9732	0.9738	0.9744	0.9750	0.9756	0.9761	0.9767
2.0	0.9772	0.9778	0.9783	0.9788	0.9793	0.9798	0.9803	0.9808	0.9812	0.9817
2.1	0.9821	0.9826	0.9830	0.9834	0.9838	0.9842	0.9846	0.9850	0.9854	0.9857
2.2	0.9861	0.9864	0.9868	0.9871	0.9875	0.9878	0.9881	0.9884	0.9887	0.9890
2.3	0.9893	0.9896	0.9898	0.9901	0.9904	0.9906	0.9909	0.9911	0.9913	0.9916
2.4	0.9918	0.9920	0.9922	0.9925	0.9927	0.9929	0.9931	0.9932	0.9934	0.9936
2.5	0.9938	0.9940	0.9941	0.9943	0.9945	0.9946	0.9948	0.9949	0.9951	0.9952
2.6	0.9953	0.9955	0.9956	0.9957	0.9959	0.9960	0.9961	0.9962	0.9963	0.9964
2.7	0.9965	0.9966	0.9967	0.9968	0.9969	0.9970	0.9971	0.9972	0.9973	0.9974

Table 1 Area Φ(x) under the standard normal curve to the left of x—cont’d
2.8	0.9974	0.9975	0.9976	0.9977	0.9977	0.9978	0.9979	0.9979	0.9980	0.9981
2.9	0.9981	0.9982	0.9982	0.9983	0.9984	0.9984	0.9985	0.9985	0.9986	0.9986
3.0	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9989	0.9989	0.9990	0.9990
3.1	0.9990	0.9991	0.9991	0.9991	0.9992	0.9992	0.9992	0.9992	0.9993	0.9993
3.2	0.9993	0.9993	0.9994	0.9994	0.9994	0.9994	0.9994	0.9995	0.9995	0.9995
3.3	0.9995	0.9995	0.9995	0.9996	0.9996	0.9996	0.9996	0.9996	0.9996	0.9997
3.4	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9998	0.9998
3.5	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998
3.6	0.9998	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999
3.7	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999
3.8	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	1.0000	1.0000	1.0000

Question 6.22: Assume that the random variable Sn is the sum of 48 independ......

Related Answered Questions

(A more general case of Example 6.3) Obtain the PDF of Z=X+Y, where X and Y are two independent random variables with the following PDFs: fX (x)= {1/b-a a<x<b 0 otherwise fY (y)= 1/d-c c<y <d, d-c<b-a 0 otherwise ...

Verified Answer:

Find the PDF of the sum of X and Y if the two random variables are independent random variables with the common PDF fX (u)=fY (u)= {1/4 0<u<4 0 otherwise ...

Verified Answer:

Find the PDF of the random variable Y=aX², where X is the standard normal random variable and a>0. ...

Verified Answer:

Find the PDF of Y in terms of the PDF of X if Y=2X+7. ...

Verified Answer:

Find the PDF of U, which is the sum of X and Y that are independent random variables with the following PDFs: fX(x) = {λe^-λx x ≥ 0 fY(y) = {λ²ye^-λy y ≥ 0 ...

Verified Answer:

The time X between consecutive snowstorms in winter is a random variable with the PDF fX(x) = {λe^-λx x ≥ 0 0 otherwise Assume it has not snowed up until now. What is the PDF of the time U until the second snowstorm? ...

Verified Answer:

Find the PDF of W, which is the sum of X and Y that are independent random variables with the following PDFs: fX(x)= λe^-λx x ≥ 0 fY(y)= μe^-μy y ≥ 0 where λ ≠ μ. ...

Verified Answer:

Assume that V = min(X, Y), where X and Y are independent random variables with the respective PDFs fX(x) = λe^-λx x ≥ 0 fY(y) = μe^-μy y ≥ 0 where λ > 0 and μ > 0. What is the PDF of V? ...

Verified Answer:

Verified Answer:

Verified Answer: